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Let $\mathrm{A, B, C}$ be three events. If the probability of occurring exactly one event out of $\mathrm{A\: and\: B}$ is $\mathrm{1-a}$ , out of $\mathrm{B \: and \: C}$ is $\mathrm{1-2 a}$ , out of $\mathrm{C \: and\: A}$ is $\mathrm{1-a}$ and that of occurring three events simultancously is $\mathrm{a^2}$ . If the probability that atleast one out of $\mathrm{A, B, C}$ will occur is greater than $\mathrm{\lambda}$ , then the value of $2050 \lambda$ must be
Option: 1
$1025$

Option: 2
$2050$

Option: 3
$1050$

Option: 4
$1025$

Answers (1)

best_answer

Given, $\mathrm{P(A)+P(B)-2 P(A \cap B)=1-a }$ ...........(i)

and $\mathrm{\quad P(B)+P(C)-2 P(B \cap C)=1-2 a }$ ............(ii)

and $\mathrm{ \quad P(C)+P(A)-2 P(C \cap A)=1-a }$ ................(iii)

and $\mathrm{ \quad P(A \cap B \cap C)=a^2 }$ ...............(iv)

$\mathrm{ \therefore P(A \cup B \cup C)=P(A)+P(B)+P(C)-P(A \cap B)-P(B \cap C) -P(C \cap A)+P(A \cap B \cap C) }$

$\mathrm{ =\frac{1}{2}\{P(A)+P(B)-2 P(A \cap B)+P(B)+P(C)\mathrm{ -2 P(B \cap C)+P(C)+P(A)-2 P(C \cap A)\} \mathrm{+P(A \cap B \cap C) }} }$

$\mathrm{ =\frac{1}{2}\{1-a+1-2 a+1-a\}+a^2 }$

{from Eqs.(i),(ii) and (iv)}

$\mathrm{ =\frac{3}{2}-2 a+a^2 }$

$\mathrm{ =(a-1)^2+\frac{1}{2} }$

$\mathrm{ >\frac{1}{2} }$

Here, $\mathrm{ \lambda=\frac{1}{2} }$

Then, $\mathrm{ 2050 \lambda=2050 \times \frac{1}{2} }$

$\mathrm{ =1025 }$

Hence option 1 is correct.

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